THE TREE PROPERTY AND THE FAILURE OF THE SINGULAR CARDINAL HYPOTHESIS AT ℵ ω 2

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1 THE TREE PROPERTY AND THE FAILURE OF THE SINGULAR CARDINAL HYPOTHESIS AT ℵ ω 2 DIMA SINAPOVA Abstract. We show that given ω many supercompact cardinals, there is a generic extension in which the tree property holds at ℵ ω2 +1 and the SCH fails at ℵ ω Introduction The tree property at κ + states that every tree with height κ + and levels of size at most κ has an unbounded branch. Equivalently, there are no κ + -Aronszajn trees. In the 1980 s Woodin asked if the failure of the Singular Cardinal Hypothesis (SCH) at ℵ ω implies the existence of an Aronszajn tree at ℵ ω+1. To motivate the question we note a few facts about Aronszajn trees. By Magidor and Shelah [9], the tree property holds at successors of singular limits of strongly compact cardinals. On the other hand, Solovay [12] showed that SCH holds above a strongly compact cardinal. Generalizing the original question, in 1989 Woodin and others asked if the failure of SCH at a singular cardinal κ of cofinality ω implies the existence of an Aronszajn tree at κ +. The first progress on Woodin s question was made by Gitik-Sharon [6], who showed the consistency of the failure of SCH at a singular cardinal κ together with the non-existence of special κ + -Aronszajn trees. They also pushed down their result to κ = ℵ ω 2. Then in 2009, Neeman [10] obtained the failure of the singular cardinal hypothesis at some large singular cardinal κ, together with the full tree property at κ +. It remained open whether this construction can be pushed down to smaller cardinals. In this paper we show that his result can indeed be obtained at smaller cardinals. In particular, we prove that it is consistent relative to large cardinals that ℵ ω 2 is a strong limit cardinal, 2 ℵ ω 2 > ℵ ω 2 +1, and the tree property holds at ℵ ω It still remains open whether an analogous result holds for ℵ ω. Theorem 1. Suppose that in V, κ n n < ω is an increasing sequence of supercompact cardinals and GCH holds. Then there is a generic extension in which: The author thanks Itay Neeman for his invaluable comments and corrections. 1

2 2 DIMA SINAPOVA (1) κ 0 = ℵ ω 2, (2) the tree property holds at ℵ ω 2 +1, (3) SCH fails at κ. The rest of the paper presents the proof of Theorem 1. In section 2 we define the forcing notion and give some basic properties about the forcing. Section 3 deals with a preservation lemma, which will be used to show the tree property. Finally in section 4 we prove that the tree property holds at ℵ ω The forcing Let κ n n < ω be an increasing sequence of supercompact cardinals. We start by using Laver s forcing to make κ 0 indestructably supercompact while maintaining GCH above κ 0. Let V be the resulting model. Denote κ = κ 0, ν = sup n κ n and µ = ν +. First we force with C, where C is the full support iterated collapse to make each κ n be the n-th successor of κ. Let H be C-generic over V. Then we force with Add(κ, ν ++ ). Let E be Add(κ, ν ++ ) generic over V [H]. Work in V [H][E]. Proposition 2. There is a normal measure on P κ (κ +ω+1 ), U, such that for each n < ω if we let U n be the projection of U to P κ (κ +n ) and j n = j Un, then there exists a (Col(κ +ω+2, < j n (κ))) Nn - generic filter, K n, over N n = Ult(V [H][E], U n ). Proof. See Gitik-Sharon [6]. Let U be a normal measure on P κ (κ +ω+1 ) given by the above proposition. For each n, let U n be the projection of U to P κ (κ +n ) and j n = j Un. Using standard reflection arguments, we choose sets X n U n for n < ω, such that for all x X n : κ x = κ x is κ +n x -supercompact. ( k n)o.t.(x κ +k ) = κ +k x. In particular, o.t.(x) = κ +n x. Fix generic filters K n n < ω as in the above proposition. We are ready to define the main forcing. Basically, we take the forcing in Gitik-Sharon [6] with collapses using the filters K n n < ω. We use the notation x y to denote that x y and x < κ y. Definition 3. Conditions in P are of the form p = d, p n n < ω, where for some integer l = lh(p) (the length of p), we have: (1) For n < l, p n = x n, c n such that: x n X n and for i < n, x i x n, if n < l 1, then c n Col(κ +ω+2 x n, < κ xn+1 ), and c l 1 Col(κ x +ω+2 l 1, < κ).

3 THE TREE PROPERTY AND NOT SCH AT ℵ ω 2 3 (2) For n l, p n = A n, C n such that: A n U n, A n X n, and x l 1 y for all y A n. C n is a function with domain A n, for y A n, C n (y) Col(κ y +ω+2, < κ), [C n ] Un K n. (3) if l > 0, then d Col(ω, κ +ω x 0 ), otherwise d Col(ω, κ). For a condition p, we will use the notation p = d p, p n n < ω, p n = x p n, c p n for n < lh(p), and p n = A p n, Cn p for n lh(p). The stem of p is h = d p, p n n < lh(p). Sometimes we will also denote the stem of p by d p, x, c, where x and c are with length lh(p), and for i < lh(p), p i = x i, c i. q = d q, q n n < ω p = d p, p n n < ω if lh(q) lh(p) and: d q d p, for all n < lh(p), x p n = x q n, c q n c p n, for lh(p) n < lh(q), x q n A p n and c q n C p n(x q n), for n lh(q), A q n A p n and for all y A q n, C q n(y) C p n(y) We say that q is a direct extension of p, denoted by q p, if q p and lh(q) = lh(p). For two stems h 1 and h 2, we say that h 1 is stronger or an extension of h 2 if there are conditions p 1 p 2 with stems h 1 and h 2 respectively. Note that any two conditions with the same stem are compatible. That is since the collapsing part is taken to be in the filters K n. For two conditions p, q with the same stem we define p q to be the weakest common extension. Let G be P generic over V [H][E], and let x n n < ω, where each x n P κ (κ +n ), be the components derived from the generic set G. Set λ n = x n κ. By Gitik-Sharon [6], we get: Proposition 4. (1) If A n n < ω V [H][E] is a sequence of sets such that every A n U n, then for all large n, x n A n. (2) n x n = ν = (κ +ω ) V [H][E]. (3) For each n 0, the cofinality of κ n = (κ +n ) V [H][E] in V [H][E][G] is ω. (4) Since any two conditions with the same stem are compatible, P has the µ = (κ +ω+1 ) V [H][E] chain condition. So, cardinals greater than or equal to µ are preserved. (5) In V [H][E][G], 2 κ = µ + and GCH holds from µ upward.

4 4 DIMA SINAPOVA (6) P has the Prikry property. I.e. if p is a condition with length at least 1 and φ is a formula, then there is a direct extension p p which decides φ. Remark 5. The main point in the proof of the Prikry property is the diagonal lemma, which states that for p P with length at least 1 and lh(p) < n < ω if H is a set of stems of length lh(p)+n and q h h H are conditions stronger than p such that each q h has a stem h, then there is q p such that if r q is a condition of length at least lh(p) + n, then r q h for some h H. Remark 6. Using the closure of the collapsing posets, we get the following corollary to the Prikry property: If p is a condition with length n + 1 and φ is a formula, then there is a direct extension q p, such that q n = p and if r p decides φ, then r n q (ω \ n) decides φ. The last property implies that all cardinals χ, such that λ n χ λ +ω+2 n for some n > 0 are preserved. In particular, in V [H][E][G] every λ n for n > 0 is a cardinal. And so, κ remains a cardinal, as well. Also, P preserves (λ +ω+1 0 ) V [H][E], (λ +ω+2 0 ) V [H][E], and collapses (λ +ω 0 ) V [H][E] to ω. So in V [H][E][G], κ = ℵ ω 2, µ = κ +, (λ +ω+1 0 ) V [H][E] = ℵ 1, and 2 ℵ ω 2 = ℵ ω So, SCH fails at ℵ ω The preservation lemma In this section we prove a preservation lemma, which will be used to show the tree property. The proof of this lemma is motivated by the Preservation Theorem in Magidor-Shelah [9]. The main difference is that instead of trees, here we are working with narrow systems, which are defined below. Throughout this section V will denote some arbitrary ground model. We say that a poset is χ-closed if it is closed under sequences of length χ. We start with defining the notion of a narrow system. This is the same as the definition given in Magidor- Shelah [9]. Definition 7. S = I κ, R is a narrow system of height ν + and levels of size κ < ν if: I is an unbounded subset of ν +, and for each α I, S α = {α} κ is the α-level of S, R is a set of transitive binary relations on I κ, such that for all R R, if α, δ R β, ξ, then α < β. R < ν. For every α < β in I, there are u S α, v S β, and R R such that u, v R.

5 THE TREE PROPERTY AND NOT SCH AT ℵ ω 2 5 For all R R, if u 1, u 2 are distinct elements of I κ such that u 1, v R and u 2, v R, then u 1, u 2 R or u 2, u 1 R. For a 1, a 2 S and R R we write a 1 R a 2 if a 1, a 2 / R and a 2, a 1 / R, and in that case say that a 1, a 2 are R-incomparable. A branch of S is a set b α I S α such that for some R R, we have that for all u, v b, u, v R or v, u R. In this case we say that b is a branch through R (or with respect to R). We say that b I κ is unbounded if for unboundedly many α I, b S α. Theorem 8. Suppose that ν is a singular cardinal of cofinality ω, κ, τ < ν are regular cardinals, and in V, Q is κ + c.c notion of forcing and R is a max(κ, τ) + closed notion of forcing. Let E be Q-generic over V and let F be R-generic over V [E]. Suppose that S = I, R is a narrow system in V [E] of height ν +, levels of size κ, and with R = R σ σ < τ. Suppose that in V [E][F ] there are (not necessarily all unbounded) branches b σ,δ σ L, δ < κ, such that: (1) every b σ,δ is a branch through R σ, and for some σ, δ L κ, b σ,δ is unbounded. (2) for all α I, there is σ, δ L κ, such that S α b σ,δ. Then S has an unbounded branch in V [E]. Proof. Let σ < τ, δ < κ. For α < ν +, p Q, and r 1, r 2 R, we say that p Q r 1, r 2 force contradictory values about ḃσ,δ S <α if there are nodes u 1 = α 1, ξ 1, u 2 = α 2, ξ 2 with α 1 < α 2 < α, such that: p Q α 1, α 2 I and u 1, u 2 / Ṙσ p Q r 1 R u 1 ḃσ,δ and r 2 R u 2 ḃσ,δ. The following lemma uses an argument from a branch lemma by Spencer Unger [13]. Versions of this lemma were also proved by Baumgartner, Foreman and Spinas while they were working on [1]. Lemma 9. (Splitting Lemma) Let σ < τ, δ < κ. Suppose that for some β < ν +, p Q r R ḃ σ,δ is an unbounded branch or r R ḃ σ,δ S <β, and p forces that Ṡ has no unbounded branches through Ṙσ in V Q. Let r 1, r 2 R be stronger than r. Then there are r 1 r 1, r 2 r 2, a maximal antichain A Q of conditions below p and α < ν + with α β, such that for all p A, p Q r 1 and r 2 force contradictory values about S <α ḃσ,δ, or p Q r R ḃ σ,δ S <β. Proof. First we show the following claim. Claim 10. For any r 1 r 1 and r 2 r 2 the set D r 1,r 2 = {p p p Q r R ḃ σ,δ S <β or ( r 1 r 1)( r 2 r 2)( α < ν + )(p Q r 1, r 2 force contradictory values for ḃσ,δ S <α )} is dense below p.

6 6 DIMA SINAPOVA Proof. Let r 1 r 1, r 2 r 2, p p. Let q p be such that q Q r R ḃ σ,δ is unbounded, or q Q r R ḃ σ,δ S <β. If the latter holds, than q D r 1,r 2, so we are done. Now suppose that q Q r R ḃ σ,δ is unbounded. Let E be Q- generic over V with q E. In V [E ] define E 1 = {a I κ ( r r 1)(r a ḃσ,δ)} and E 2 = {a I κ ( r r 2)(r a ḃσ,δ)}. Then since in V [E ], r forces that ḃσ,δ is unbounded, we have that both {α I E 1 S α } and {α I E 2 S α } are unbounded in I. Also since S has no unbounded branches through R σ in V [E ], we have that for any a E 1, there are R σ incomparable nodes c, d in E 1 with a, c R σ and a, d R σ. The same holds for E 2. So, we can find R σ incomparable nodes c 1 E 1 and c 2 E 2 as follows. Let a 1, a 2 be R σ -incomparable nodes in E 1 and let c 2 E 2 be of level above the levels of a 1, a 2. Then c 2 must be R σ -incomparable with at least one of these nodes. Let c 1 be a 1 if a 1 Rσ c 2, otherwise let c 1 = a 2. Then there are p q, r 1 r 1, r 2 r 2, and α < ν +, such that p Q r 1, r 2 force contradictory values for ḃσ,δ S <α, as witnessed by c 1 and c 2. The rest of the proof of the lemma uses Spencer Unger s diagonal construction argument to get the antichain and r1, r2 as desired. The only difference is that here all antichains are below p, and r1 0 r 1, r2 0 r 2. We include it for completeness. Inductively construct antichains A ξ ξ < κ + in Q, conditions r1, ξ r ξ 2 ξ < κ + in R and α ξ ξ < κ + as follows. Let A 0 =, α 0 = β, r1 0 = r 1, and r2 0 = r 2. For δ limit, set A δ = ξ<δ A ξ and r1, δ r2 δ to be lower bounds of r ξ 1 ξ < δ and r ξ 2 ξ < δ, respectively. Also set α δ = sup ξ<δ α ξ. At successor stages, suppose we have defined A ξ. If A ξ is a maximal antichain below p, then set A ξ+1 = A ξ, α ξ+1 = α ξ, r ξ+1 1 = r ξ 1 and r ξ+1 2 = r2. ξ Otherwise, let q p be such that q is incompatible with every condition in A ξ. Applying the claim, let q q be such that q D rξ 1,rξ 2. If q Q r R ḃ σ,δ S <β, set α ξ+1 = α ξ, r ξ+1 1 = r1, ξ and r ξ+1 2 = r2. ξ Otherwise let α ξ+1 > α ξ, r ξ+1 1 r1, ξ and r ξ+1 2 r ξ 2 be such that q Q r ξ+1 1, r ξ+1 2 force contradictory values for ḃ σ,δ S <αξ+1. Set A ξ+1 = A ξ {q }. Since Q has the κ + chain condition, for some δ < κ +, we have that A δ is a maximal antichain below p. Set α = α δ, r1 = r1 δ and r2 = r2. δ Then A δ, α, r1, and r2 are as desired. Let λ = max(κ, τ) +. The Splitting lemma yields the following corollary.

7 THE TREE PROPERTY AND NOT SCH AT ℵ ω 2 7 Corollary 11. Let σ < τ, δ < κ. Suppose that for some β < ν +, p Q r R ḃ σ,δ is an unbounded branch or r R ḃ σ,δ S <β, and p forces that Ṡ has no unbounded branches through Ṙσ in V Q. Let r i i < λ be conditions in R that are stronger than r. Then there is a sequence r i i < λ and α < ν +, such that each r i r i and for all i < j < λ, there is a maximal antichain A ij Q of conditions below p, such that for all p A ij, p Q r i and r j force contradictory values about S <α ḃσ,δ, or p Q r R ḃ σ,δ S <β. Proof. By induction on η < λ, we build sequences r η i i < η η < λ of conditions in R, such that: (1) for all i < λ, r η i η > i is decreasing, (2) for all i < j < λ, there is a maximal antichain A ij in Q of conditions below p and α ij < ν +, such that for all p A ij, p Q r j+1 i, r j+1 j force contradictory values about ḃσ,δ S <αij or p Q r R ḃ σ,δ S <β. Set r0 1 = r 0. At limit stages, suppose ρ < λ is limit and we have defined r η i i < η η < ρ. For i < ρ, set r ρ i to be stronger than each r η i for i < η < ρ. For successor stages, suppose we have defined r ρ i i < ρ. Define r ρ+1 i i ρ as follows. Inductively apply the splitting lemma for all i < ρ to get a decreasing sequences of conditions q i i < ρ in R below r ρ and r ρ+1 i r ρ i, such that for every i < ρ, there is α iρ < ν + and an antichain A iρ of conditions below p, such that for all p A iρ, p Q r ρ+1 i, q i force contradictory values about ḃσ,δ S <αiρ or p Q r R ḃ σ,δ S <β. Then set rρ ρ+1 to be stronger than each q i. This completes the construction. Finally let r i be stronger than each r η i for i < η < λ, and α = sup i<j<λ α ij. We return to the proof of the theorem. Suppose for contradiction that S has no unbounded branch in V [E]. Since R is max(κ, τ) + distributive in V [E], we can find r F and β 0 < ν +, such that for all σ, δ τ κ, either r ḃσ,δ is unbounded, or r ḃσ,δ S <β0. Also by (2) from the assumptions of the theorem, by further strengthening r, we can assume: ( ) r for all α I, there is σ, δ such that S α ḃσ,δ. Now let p E be such that p forces that for all σ, δ τ κ, either r ḃσ,δ is unbounded, or r ḃσ,δ S β0. Suppose also

8 8 DIMA SINAPOVA that p forces that Ṡ has no unbounded branches in V Q. Working in V, we will define a sequence r i i < λ of conditions stronger than r as follows. For every σ, δ τ κ using the last corollary we build conditions r σ,δ i i < λ stronger than r and β σ,δ < ν +, such that: (1) for every i, r σ,δ i σ, δ τ κ is decreasing according to some enumeration of τ κ, (2) for every σ, δ τ κ and every i < j < λ, there is a maximal antichain A σ,δ i,j in Q of conditions below p, such that, for every p A σ,δ ij, p Q r σ,δ i and r σ,δ j force contradictory values about ḃ σ,δ S <β σ,δ or p Q r R ḃ σ,δ S <β0. Then for every i < λ, set r i to be stronger than all of r σ,δ i σ, δ τ κ. Work in V [E]. Let β I be such that β > sup σ,δ β σ,δ. For all i < λ, let r i r i be such that for some ξ i, σ i, δ i, r i β, ξ i ḃσ i,δ i. We can find such r i by ( ). Since λ > max(τ, κ), for some ξ < κ and σ, δ τ κ, there are distinct i < j < λ, such that ξ i = ξ j = ξ, σ i = σ j = σ and δ i = δ j = δ. Let p E A σ,δ ij. Then there is p p, such that: (1) p Q r i, r j force contradictory information about ḃσ,δ S <β or p Q r forces that ḃσ,δ S <β0, (2) p Q β I, ṙ i r i, ṙ j r j, (3) p Q ṙ i forces that β, ξ ḃσ,δ, (4) p Q ṙ j forces that β, ξ ḃσ,δ If p Q r forces that ḃσ,δ S <β0, then by (2), we get that p Q ṙ i forces that ḃσ,δ S <β0, but this is a contradiction with (3). On the other hand if p Q r i, r j force contradictory information about ḃσ,δ S <β, then by (2), it follows that p Q ṙ i, ṙ j force contradictory information about ḃσ,δ S <β. But since ḃσ,δ is forced to be a branch, then by (3) and (4) we have that p must force that ṙ i and ṙ j force the same values for ḃσ,δ at levels below β. Contradiction. Remark 12. We can actually get something stronger. Starting from the assumptions of the Preservation Theorem, we can show that for some σ < τ and δ < κ, b σ,δ is unbounded, and there is an unbounded branch b in V [E] through R σ with b σ,δ b. To do this, we just modify the assumptions of the Splitting Lemma to state that p forces that either

9 THE TREE PROPERTY AND NOT SCH AT ℵ ω 2 9 r ḃσ,δ S <β or r ḃσ,δ is unbounded and Ṡ has no unbounded branch b in V Q through Ṙσ, such that ḃσ,δ b. Then carry over the same assumption to the corollary to the splitting lemma. The rest of the argument is exactly the same. We will use this stronger version in Lemma 16 of next section. 4. The tree property In this section we will show that in V [H][E][G] the tree property holds at κ +. Let T in V [H][E] be a P - name for a ν + tree with levels of size at most κ, such that this is forced by the empty condition. Denote the α-th level of T by T α. We may assume that T α = {α} κ for α < ν +. We will show that T has a cofinal branch in V [H][E][G]. The outline of our proof is motivated by Neeman [10]. The main difference is that we have to deal with the poset C and rely on the Preservation Lemma from the last section. Lemma 13. There is n < ω and an unbounded I ν + in V [H][E], such that for all α < β in I, there are ξ, δ < κ and a condition q with length n, such that q α, ξ < T β, δ. Proof. Recall that U was the normal measure on P κ (ν + ) fixed in advance and each U n is the projection of U to P κ (κ n ). Let j = j U : V [H][E] M. Let G be j(p) - generic over M and T = j( T ) G be such that the first element of the generic sequence added by G is κ. We can arrange that since κ j(x 0 ). Then (ν + ) V remains a regular cardinal in M[G ]. Fix a node u T of level γ, where sup(j ν + ) < γ < j(ν + ). Then for all α < ν + let ξ α < j(κ) be such that j(α), ξ α < T u and p α G be such that p α j(α), ξ α < j( T ) u. Then in M[G ] there is an unbounded I ν + and a fixed n, such that for all α I, p α has length n. Denoting p α = d α, p α i i < ω, by further shrinking I we can assume that for some d Col(ω, κ +ω ), for each α I, d α = d. Also, for each α I and i < n, denote p α i = y i, c α i. Note that by choice of G, we have that y 0 = κ. Let b = d, y, c be a stem in j(p) with length n such that y = y i i < n and c = c i i < n where each c i = α cα i. We can take this union since for 0 < i < n, c i belongs to a poset which is < (j(κ) y i ) +ω+2 closed, and c 0 Col(κ +ω+2, < j(κ) y1 ). In particular, the closure is larger than ν + = κ +ω+1. Define I = {α < ν + p j(p) stem(p) = b, and ξ < j(κ)p j(α), ξ < j( T ) u}. Then I V [H][E] and I I, so I is unbounded. So, I is as desired.

10 10 DIMA SINAPOVA Remark 14. Note that for any condition p, we can strengthen Lemma 13 to get an unbounded set I and n < ω, such that for all α < β in I, there are ξ, δ < κ and a condition q p with length n, such that q α, ξ < T β, δ. To do this, if k = lh(p), we choose G, so that the k th element of the generic sequence added by G below j(κ) is κ and below that we take the stem of p. More precisely, if x i i < ω is the G -generic sequence, we arrange so that j(κ) x k = κ, and for i < k, x i = j(x p i ) = j xp i. Let n and I be as in the conclusion of the above lemma. We will say that a stem h φ if there is a condition p, such that the stem of p is h and p φ. Since any two conditions with the same stem are compatible, we have that if h φ, then h φ Lemma 15. There is, in V [H][E], an unbounded set J ν +, a stem h of length n, and a sequence of nodes u α α J with every u α of level α, such that for all α < β in J there is a condition p with stem h, such that p u α < Proof. Let j : V N be a ν + - supercompact embedding with critical point κ n+2. Using standard arguments, extend j to j : V [H] N[H ], where H is j(c) generic over V. We can arrange so that H = H E H, where H is C - generic over V [H][E] (see [8]). We have that C is < κ n+1 distributive in V [H][E], and it is < κ n+1 closed in V [H]. Let F be Add(κ, j(ν ++ )) generic over V [H][E]. Then F is also generic over V [H][E][H ], since Add(κ, j(ν ++ )) has the κ + chain condition. Define E = {f : j(ν ++ ) κ κ f < κ, f (j ν ++ ) j E, f (j(ν ++ ) \ j ν ++ ) F }. Then E is Add(κ, j(ν ++ )) generic over V [H ] = V [H][E][H ], such that j E E. It follows that we can extend j to j : V [H][E] N[H ][E ] where j V [H][E][F ][H ] = V [H][E][H ][F ]. Let γ j (I) be such that sup(j ν + ) < γ < j(ν + ). By elementarity for all α I we can fix ξ α, δ α < κ and p α j (P) with length n such that p α j (P) j (α), ξ α < j ( T ) γ, δ α. Let h α be the stem of p α. Note that the function α ξ α, δ α, h α is in V [H][E][F ][H ]. First we will use the Preservation Lemma to show that J as in the statement of the lemma exists in V [H][E][F ]. Then we will use the arguments in Neeman [10] to get J V [H][E]. The number of possible stems in j (P) of length n is less than κ n. Then since ν + is regular in V [H][E][F ] and C does not add sequences of length less than κ n, we have that in V [H][E][F ][H ] there is a cofinal J I, δ < κ, and a stem h such that for all α J, δ α = δ and h α = h.

11 THE TREE PROPERTY AND NOT SCH AT ℵ ω 2 11 We consider the narrow system S = I, R of height ν + and levels of size κ, in V [H][E] (and so in V [H][E][F ]), where: R = R h h is a stem of length n ; R < κ n. For nodes a, b, we say that a, b R h iff h a < T b Note that {h h is a stem of length n} V [H]. Also each R h is transitive since conditions with the same stem are compatible and T is forced to be a tree by the empty condition. Apply the preservation lemma to S for V 1 = V [H], Q = Add(κ, j(ν ++ )), and R = C, and the branches: b h,δ = def { α, ξ h j (P) j(α), ξ < j ( T ) γ, δ }. Note that the preservation lemma works for any R = R σ σ L with the index set L in the ground model for the lemma, which in this case is V 1. We get that S has an unbounded branch in V [H][E][F ]. I.e. in V [H][E][F ], there are an unbounded J I, α ξ α and a stem h such that for all α, β J with α < β, we have that h α, ξ α < T β, ξ β. Then by the argument in Lemma 3.2 from [10], we can get such J, h, α ξ α in V [H][E]. Setting u α = α, ξ α for α J, we get that for all α < β in J there is a condition p with stem h which forces that u α < Fix n, h, J, and α u α as in the conclusion of the above lemma. By shrinking J we may assume that for some ξ < κ, each u α = α, ξ. Lemma 16. Suppose that h is a stem of length k, L ν + is unbounded, and for all α < β with α, β L, h u α < Then there are ρ < ν + and sets A α, C α : α L \ ρ in V [H][E] such that: (1) Each A α U k, dom(c α ) = A α, C α (x) Col(κ +ω+2 x, < κ) for x A α, and [C α ] Uk K k. (2) For all α < β in L \ ρ, for all x A α A β such that C α (x) and C β (x) are compatible, h x, C α (x) C β (x) u α < T u β. Proof. Let j : V N be a ν + - supercompact embedding with critical point κ k+3. As in the previous lemma, extend j to j : V [H] N[H ], where H = H E H is j(c) generic over V, and H is C - generic over V [H][E]. We have that C is < κ k+2 distributive in V [H][E] and < κ k+2 closed in V [H]. Let F be Add(κ, j(ν ++ )) generic over V [H][E]. Then F is also generic over V [H][E][H ], and so E = def {f : j(ν ++ ) κ κ f < κ, f (j ν ++ ) j E, f (j(ν ++ ) \ j ν ++ ) F } is Add(κ, j(ν ++ )) generic over V [H ] = V [H][E][H ], such that j E

12 12 DIMA SINAPOVA E. It follows that we can extend j to j : V [H][E] N[H ][E ] where j V [H][E][F ][H ] = V [H][E][H ][F ]. First we will adapt the arguments in Neeman [10] to include the collapses and use them to show that there are ρ < ν +, α A α and α Cα as in the statement of the lemma in V [H][E][H ]. Then we will use the Preservation Lemma to show that we can find these objects in V [H][E]. For all α < β in L, by the assumptions of the lemma, we can fix (in V [H][E]) [C α,β ] Uk K k, such that for almost all x P κ (κ k ), h x, C αβ (x) u α < Since K k is closed under sequences of length κ +ω+1, let [C ] Uk K k be stronger than [C α,β ] Uk for all α < β in L. Then for all α < β in L, there is r P with stem h, such that r u α < T u β, and denoting r k = A r k, Cr k, we have that [Ck r] U k = [C ] Uk. Without loss of generality we can assume that dom(c ) = P κ (κ k ). Claim 17. In V [H][E][H ], there are ρ < ν +, α A α and α C α that satisfy the conclusion of the lemma. Proof. The argument follows closely Lemma 3.5 in [10], so we only outline the main points. The difference here is that we are also dealing with collapses. Let γ j (L) be such that j ν + < γ < j(ν + ). We write u γ for the γ th member of the sequence j ( u α : α L ). By elementarity of j, we can find conditions r α : α L in N[H ][E ], such that each r α j (P), the stem of r α is h and r α j (P) j(u α ) < j ( T ) u γ. We may assume that for each α, [C rα k ] j (U k ) = j ([C ] Uk ) = [C ] j (U k ). For x P κ (κ k ), set J x = {α L h x, C (x) j(u α ) < j ( T ) u γ}. Also, let K x = {C L C is unbounded and ( b Add(κ, j(ν ++ )))(b J x = C)}. Then each K x and x K x is in V [H][E][H ]. We have that (see [10]): If J x is unbounded in ν +, then J x V [H][E][H ]. If C 1 C 2 are both in K x, then they are disjoint on a tail end. For all C K x, if α < β in C, then h x, C (x) u α < Then let ρ < ν + be such that for all x and C 1, C 2 K x, C 1 and C 2 are disjoint above ρ. For α L \ ρ and x P κ (κ k ) define f(x, α) to be the unique C K x such that α C if such a C exists and undefined otherwise. Note that if J x is unbounded in ν + and α J x, then f(x, α) = J x. Let α 0 = min(l \ ρ) and for α L \ ρ, set: A α = {x P κ (κ k ) f(x, α 0 ) = f(x, α)}.

13 THE TREE PROPERTY AND NOT SCH AT ℵ ω 2 13 By the arguments in [10] and since for each α, [C rα k ] = [C ], each A α U k. We remark that here when adapting the argument from [10], we use that H adds no sequences of length κ k and that ν + is regular in V [H][E]. Set Cα = C A α. Then ρ, α A α and α Cα are as desired. Fix ρ < ν +, α A α and α Cα as in the above claim. Note that for each α, Cα = C A α. Let L = L \ ρ. For every x P κ (κ k ), let b x = {α L \ ρ x A α}. Note that each b x V [H][E][H ] and for all α < β in b x, h x, C (x) u α < Claim 18. For every A U k, there is x A, such that b x is unbounded and there is an unbounded set b ν + in V [H][E], such that: b x b, for all α < β in b, h x, C (x) u α < Proof. Let R = {R x x P κ (κ k )}; R = κ k. We consider the narrow system S = L, R, where for α L, S α = {α} κ and for nodes a, b, we say that a, b R x iff: h x, C (x) u α < T u β Each R x is a transitive relation since T is forced to be a tree by the empty condition and any two conditions with the same stem are compatible. For x P κ (κ k ), set b x = {u α α b x } if x A, and b x = otherwise. We have that {b x x P κ (κ k )} are branches in V [H][E][H ], such that: (1) each b x is a branch through R x. (2) for all α L, there is x, such that b x S α (just take any x A α A), (3) for some x, b x is unbounded. (3) holds because otherwise, for all x, let α x be a bound for b x (in V [H][E][H ]) and let α = sup x Pκ(κk ) α x. Since H does not add sequences of length κ k+1 and ν + is regular in V [H][E], we have that α < ν + ; contradiction. We apply the preservation lemma to S for V 1 = V [H], Q = Add(κ, ν ++ ), R = C, and these branches to get an unbounded branch through S in V [H][E]. So, in V [H][E] we have an unbounded b L and x P κ (κ k ), such that b x b and b x is unbounded. Here we use the stronger version of the preservation lemma, see Remark 12. Since b x is unbounded, we know that x A. So, since the preservation lemma produces a branch through R x, we have that for all α < β in b, h x, C (x) u α <

14 14 DIMA SINAPOVA For every x P κ (κ k ), let x be the statement: b x is unbounded, and there is an unbounded set b ν + in V [H][E] with b x b, such that for all α < β in b, h x, C (x) u α < Corollary 19. A = def {x P κ (κ k ) x holds } V [H][E] and A U k. Proof. A is in V [H][E] since C is distributive enough. Now suppose for contradiction that A is not measure one. Then Y = {x x does not hold } U k. Apply the above claim to Y to get a contradiction. For every x A, let L x L witness x. I.e. L x V [H][E], it is unbounded in ν +, b x L x, b x is unbounded, and for all α < β in L x, we have that h x, C (x) u α < Note that by distributivity of C, we have that x L x is in V [H][E]. Finally, for α L, define: A α = {x A α L x }. Claim 20. For all α L, A α U k Proof. Suppose otherwise. Then Y = P κ (κ k ) \ A α U k. Let x Y A α A. Then x holds, and α b x L x. But we assumed that x / A α, contradiction For α L, define C α = C A α. Now suppose α < β in L, x A α A β and c C α (x) C β (x) = C (x). Then α, β L x, and so h x, C (x) u α < So, h x, c u α < Lemma 21. There is some ρ < ν + and a sequence of conditions p α α J \ρ with stem h such that for α < β in J \ρ, p α p β u α < (Recall that p α p β denotes the weakest common extension of p α and p β.) Proof. The proof is an adaptation of the argument given in [10]. First we make some remarks on taking diagonal intersections. Let H be a set of stems of length n, and let A h h H be a sequence of U n - measure one sets. For a stem h = d, y, c in H and z P κ (κ n ), we write h z to denote that y n 1 z, i.e. that y n 1 < κ z and y n 1 z. Note that h z iff for some c, h z, c is a stem. Then A = h H A h = {z P κ (κ n ) z h z Ah } is the diagonal intersection of A h h H and A U n. We will define sequences ρ n n n < ω, A n α α J \ ρ n, n n < ω, and C n α α J \ ρ n, n n < ω by induction on n, such that for all n:

15 THE TREE PROPERTY AND NOT SCH AT ℵ ω 2 15 (1) For all α J \ ρ n, we have that A n α U n, [Cα] n Un K n, dom(cα) n = A α, and Cα(x) n Col(κ +ω+2 x, < κ) for x A n α. (2) For all α < β in J \ ρ n, for all stems h = d, x, c of length n + 1 extending h, such that for n i n, x i A i α A i β and c i Cα(x i i ) Cβ i (x i), h u α < T u β. Let ρ n and A n α, Cα n α J \ρ n be given by the above lemma applied to h. Then by the conclusion of the lemma, both of the conditions above for n are satisfied. Now suppose we have defined ρ n and A n α, Cα n α J \ ρ n such that (1) and (2) above hold for n. We have to define ρ n+1 and A n+1 α, Cα n+1 for α J \ ρ n+1. For a stem h = d h, x, c of length n + 1 extending h, we say that h fits α iff each x i A i α and c i Cα(x i i ) for i n. Set J h = {α J \ ρ n h fits α}. Define a function h ρ h on stems of length n + 1 extending h as follows: if J h is bounded in ν +, let ρ h < ν + be a bound, otherwise, let ρ h and A h α, C h α α J h \ ρ h be given by the previous lemma applied to h and J h (here we use the inductive assumption for n). Set ρ n+1 = sup{ρ h h is a stem of length n + 1 extending h}. For α J \ ρ n+1, set H α (n + 1) = {h h has length n + 1, extends h, and fits α}. For each α J \ ρ n+1, let A n+1 α = h Hα(n+1)A h α. Also set [C α ] Un+1 = {[Cα] h Un+1 h H α (n+1)} K n+1. By shrinking A n+1 α (via diagonal intersections), we can arrange that for all x A n+1 α, C α (x) = {Cα(x) h h x}. It remains to check that (1) and (2) hold for n+1. The first condition holds by construction. For the second condition, we have to show that for all α < β in J \ ρ n+1, for all stems t of length n + 2, which extend h and fit both α and β, we have that t u α < T u β. So, suppose that α < β are in J \ ρ n+1 and h x, c is a stem as above. Since h fits α and β and ρ h < α, β, we have that J h is unbounded and ρ h was given by applying the previous lemma. Then h H α (n + 1) H β (n + 1) and since we took diagonal intersections, it follows that x A h α A h β and c Ch α(x) Cβ h (x). So, by construction of

16 16 DIMA SINAPOVA A h α, A h β, Ch α, Cβ h, we have that h x, c u α < This completes the construction. Now let ρ = sup n ρ n and set p α for α J \ ρ to be such that: the stem of p α is h, for all n n, let p α n = A n α, Cα. n Then p α α J \ ρ is as desired. For if α < β are in J \ ρ, and q p α p β, then by the construction, we have that stem(q) u α < T u β, and so q u α It follows that p α p β u α < Lastly, we show that {u α α < ν +, p α G} is an unbounded branch of T. It suffices to prove the following: Proposition 22. B = {α < ν + p α G} is unbounded. Proof. Otherwise, let q G be such that q Ḃ is bounded. Since both Lemma 13 and Lemma 15 can be done below any condition, we may assume that (by strengthening q if necessary) stem(q) = h. P has the ν + chain condition, so for some α < ν +, q Ḃ α. Let β J \ α, and let r be a common extension of q and p β. Then on one hand we have that r p β Ġ, but also r u β Ḃ. Contradiction. Then {u α α B} is an unbounded branch of T. This completes the proof of the tree property. To summarize, starting from ω many supercompact cardinals, we have constructed a generic extension in which there are no Aronszajn trees at ℵ ω 2 +1 and the SCH fails at ℵ ω 2. We conclude this paper with an open problem. It is still not known whether it is consistent to have the tree property at ℵ ω+1 and not SCH at ℵ ω. References [1] James Baumgartner, Matthew Foreman, Otmar Spinas, The spectrum of the Γ-invariant of a bilinear space., J. Algebra, vol. 189(1997), no. 2, pp [2] James Cummings and Matthew Foreman, Diagonal prikry extensions, J. of Symbolic Logic, vol. 75(2010), no. 4, pp [3] James Cummings, Matthew Foreman and Menachem Magidor, Squares, scales and stationary reflection, J. of Math. Log., vol. 1(2001), pp [4], Canonical structure in the universe of set theory I, Ann. Pure Appl. Logic, vol. 129(2004), no. 1-3, pp [5], Canonical structure in the universe of set theory II, Ann. Pure Appl. Logic, vol. 142(2006), no. 1-3, pp [6] Moti Gitik and Assaf Sharon, On SCH and the approachability property, Proc. of the AMS, vol. 136 (2008), pp

17 THE TREE PROPERTY AND NOT SCH AT ℵ ω 2 17 [7] Thomas Jech, Set theory, Springer Monographs in Mathematics, Springer- Verlag, [8] Menachem Magidor, Reflecting Stationary Sets., J. of Symbolic Logic, vol. 47 (1982), no. 4, pp [9] Menachem Magidor and Saharon Shelah, The tree property at successors of singular cardinals., Arch. Math. Logic, vol. 35 (1996), no. 5-6, pp [10] Itay Neeman, Aronszajn trees and the failure of the singular cardinal hypothesis, J. of Mathematical Logic vol. 9 (2009, published 2010) pp [11] Saharon Shelah, Cardinal arithmetic., Oxford Logic Guides, 29, Oxford Uniersity Press [12] Robert M. Solovay, Strongly compact cardinals and the GCH, Proceedings of the tarski symposium (proc. sympos. pure math., vol. xxv, univ. california, berkeley, calif., 1971), Amer. Math. Soc., , [13] Spencer Unger, preprint Department of Mathematics, University of California Irvine, Irvine, CA , U.S.A address: dsinapov@math.uci.edu